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Find the sine of the angle between the vectors aijka→=3i^+j^+2k^ and bijkb→=2i^-2j^+4k^. - Mathematics

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प्रश्न

Find the sine of the angle between the vectors `vec"a" = 3hat"i" + hat"j" + 2hat"k"` and `vec"b" = 2hat"i" - 2hat"j" + 4hat"k"`.

योग
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उत्तर

Given that `vec"a" = 3hat"i" + hat"j" + 2hat"k"` and `vec"b" = 2hat"i" - 2hat"j" + 4hat"k"`.

We know that `|vec"a" xx vec"b"| = |vec"a"||vec"b"| sin theta`

∴ `vec"a" xx vec"b" = |(hat"i", hat"j", hat"k"),(3, 1, 2),(2, -2, 4)|`

= `hat"i"(4 + 4) - hat"j"(12 - 4) + hat"k"(-6 - 2)`

= `8hat"i" - 8hat"j" - 8hat"k"`

`|vec"a" xx vec"b"| = sqrt((8)^2 + (-8)^2 + (-8)^2)`

= `sqrt(64 + 64 + 64)`

= `sqrt(192)`

= `sqrt(64 xx 3)`

= `8sqrt(3)`

`|vec"a"| = sqrt((3)^2 + (1)^2 + (2)^2)`

= `sqrt(9 + 1 + 4)`

= `sqrt(14)`

`|vec"b"| = sqrt((2)^2 + (-2)^2 + (4)^2)`

= `sqrt(4 + 4 + 16)`

= `sqrt(24)`

= `2sqrt(6)`

∴ `sin theta = |vec"a" xx vec"b"|/(|vec"a"||vec"b"|)`

= `(8sqrt(3))/(sqrt(14) * 2sqrt(6))`

⇒ `(4sqrt(3))/sqrt(84) = (4sqrt(3))/(2sqrt(21))`

= `2/sqrt(7)`

Hence, `sin theta = 2/sqrt(7)`.

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अध्याय 10: Vector Algebra - Exercise [पृष्ठ २१५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 10 Vector Algebra
Exercise | Q 11 | पृष्ठ २१५

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